SOME REMARKS ON THE SYMPLECTIC GROUP Sp(2g, Z). KUMAR BALASUBRAMANIAN; GANESH JI OMAR Abstract Let G = Sp(2g, Z) be the symplectic group over the integers. Given m ∈ N, it is natural to ask if there exists a non-trivial matrix A ∈ G such that Am = I, where I is the identity matrix in G. In this paper, we determine the possible values of m ∈ N for which the above problem has a solution. We also show that there is an upper bound on the maximal order of an element in G. As an illustration, we apply our results to the group Sp(4, Z) and determine the possible orders of elements in it. Finally, we use a presentation of Sp(4, Z) to identify some finite order elements and do explicit computations using the presentation to verify their orders. 1. Introduction Given a group G and a positive integer m ∈ N, it is natural to ask if there exists k 6= e ∈ G such that km = e where e is the identity element in G. In this paper, we address this question in the case of the symplectic group Sp(2g, Z). One of the principal reasons for focussing on the group Sp(2g, Z) is the following. It plays an important role in geometry and number theory and comes up in many interesting situations. For example, in geometry it plays a significant role in the study of certain type of surfaces due to its connections with the mapping class group. It also comes up in number theory in the arXiv:1308.4934v1 [math.GR] 22 Aug 2013 study of Siegel modular forms. Before we state the main results, we first recall the definition of Sp(2g, Z) and fix some notation. The group Sp(2g, Z) is the group of all 2g × 2g matrices with integral entries satisfying ⊤ A JA = I ⊤ 0 I where A is the transpose of the matrix A and J = g g . Ç−Ig 0gå Key words and phrases. Symplectic group, Euler’s φ function. 1 α1 αk Throughout we write m = p1 ...pk , where pi is a prime and αi > 0 for all i ∈ {1, 2,...,k}. We also assume that the primes pi are such that pi <pi+1 for 1 ≤ i < k+1. Also for A ∈ G we let o(A) denote the order of A. We now state the main results of this paper. Theorem 1.1. Let A = {m ∈ N | pi ≤ 2g + 1 for some i}. For A ∈ G, we have Am = I if and only if m ∈A. Theorem 1.2. Let A ∈ G be such that o(A)= m. Then m ≤ max{30, M} g (g+1) log α α 2 where M = max{2(2g) , (2g) } with α = log3 . The paper is organized as follows. In section 2, we recall some important results that we need in the sequel. Section 3, contains the proofs of the main results (Theorem 1.1, Theorem 1.2 ) of this paper. In section 4, we explicitly identify some finite order elements in G when g = 2. 2. Preliminaries Throughout this section we let φ denote the Euler’s phi function. For the sake of completeness, we recall the definition of φ and record a few properties we need. For a detailed account of the properties of the Euler’s φ function, we refer the reader to [3]. For n ∈ N, φ(n) is defined to be the number of positive integers less than or equal to n and relatively prime to n. The function φ is a multiplicative function. i.e., for m,n ∈ N which are relatively prime, we have φ(mn) = φ(m)φ(n). Using the fact that every positive integer n> 1, can be expressed in a unique way as k n = pα1 pα2 ...pαk = pαi 1 2 k Y i i=1 where p1 < p2 < · · · < pk are primes and αi’s are positive integers and the α1 α2 αk fact that φ is multiplicative, it is clear that φ(n)= φ(p1 )φ(p2 ) ...φ(pk ). It is therefore useful to know the value of φ for prime powers. It is easy to see that − φ(pα)= pα 1(p − 1), where p is a prime and α is a positive integer. We now state the results we shall use to prove the main theorems in this paper. We refer the reader [4] and [2] for a more detailed account of these results. Theorem 2.1 (Shapiro). Let m ∈ N be such that m 6∈ {1, 2, 3, 4, 6, 10, 12, 18, 30} log2 α and α = log3 . Then φ(m) >m . 2 α1 αk Theorem 2.2 (B¨urgisser). Let m = p1 ...pk , where the primes pi satisfy pi < pi+1 for 1 ≤ i < k and where αi ≥ 1 for 1 ≤ i ≤ k. There exists a matrix A ∈ Sp(2g, Z) of order m if and only if k a) φ(pαi ) ≤ 2g, if m ≡ 2(mod4). X i i=2 k b) φ(pαi ) ≤ 2g, if m 6≡ 2(mod4). X i i=1 3. Main Results Throughout this section we take m ∈ N to be as in Theorem 2.2. We let A = {m ∈ N | pi ≤ 2g + 1 for some i} and B = N \ A. Before we prove the main theorem, we record a lemma we need. Lemma 3.1. Let A ∈ G such that o(A) = m. Then pi ≤ 2g + 1, ∀i ∈ {1, 2,...,k}. Proof. Suppose pi > 2g +1 for some i ∈ {1, 2,...,k}. This would imply that − αi αi 1 φ(pi )= pi (pi − 1) > 2g and we get a contradiction to Theorem 2.2. Theorem 3.2. Let m ∈ N. Then Am = I if and only if m ∈A. Proof. Suppose m ∈A. Choose pi such that pi ≤ 2g +1. Clearly, φ(pi) ≤ 2g and it follows from theorem 2.2 that we have A ∈ G such that o(A) = pi. − α1 αi 1 αk m pi n Let n = p1 ...pi ...pk . Now A = (A ) = I. Suppose there exists A ∈ G such that Am = I. We show that m 6∈ B. m r1 rℓ Since A = I, it follows that o(A)|m. Let o(A) = n = q1 ...qℓ . By lemma 3.1, we know that each qi ≤ 2g + 1. The result now follows from the following simple observation. Since qi|n and n|m, we have qi|m for each i ∈ {1, 2,...,ℓ}. This is indeed not possible if m ∈B. 3.1. An upper bound for the order. We show that the maximal order in G is always bounded. First, we introduce some notation. N α2 αk Let m ∈ , be as in Theorem 2.1 and let n = p2 ...pk . Suppose that there exists A ∈ G such that o(A)= m. Consider the following sums: k S = φ(pαi ) if m 6≡ 2(mod 4) and 1 X i i=1 k S = φ(pαi ) if m ≡ 2(mod 4). 2 X i i=2 In the following lemmas, We show that S1 and S2 are bounded below. 3 k α α Lemma 3.3. S = φ(p i ) >m (g+1) . 1 X i i=1 Proof. We know that pi ≤ 2g + 1, 1 ≤ i ≤ k. From this it follows that k ≤ g + 1. Now consider the sum S1. We have α α α α 1 1 k 1 k k φ(p1 )+ · · · + φ(pk ) ≥ k(φ(p1 ) ...φ(pk )) 1 = k(φ(m)) k α >m k α ≥ m (g+1) . k α α Lemma 3.4. S = φ(p i ) >n g . 2 X i i=2 α2 αk Proof. Since m ≡ 2(mod 4), it follows that m = 2n = 2(p2 ...pk ). Clearly, n 6∈ {1, 2, 3, 4, 6, 10, 12, 18, 30} and the inequality in theorem 2.1 applies. Applying a similar argument as in lemma 3.3 to n gives us the desired lower bound for S2. Theorem 3.5. Let A ∈ G be such that o(A)= m. Then m ≤ max{30, M} g (g+1) log α α 2 where M = max{2(2g) , (2g) } with α = log3 . α g Proof. Suppose that m 6≡ 2(mod 4). By lemma 3.3, we have S1 >m ( +1) . If (g+1) m> (2g) α , then we have S1 > 2g. This is clearly not possible. Thus it (g+1) follows that m ≤ (2g) α . Similarly, we see that if m ≡ 2(mod 4), then lemma 3.4 applies and we α g have S2 > n g . If m > 2(2g) α , then S2 > 2g. As this is not possible, it α follows that m ≤ 2(2g) g . g (g+1) Taking M = max{2(2g) α , (2g) α }, we obtain m ≤ max{30, M}. 4. Finite order elements in Sp(4, Z) Using Lemma 3.1, Theorem 3.5 and Theorem 2.2, it is easy to see that the possible orders of elements in Sp(4, Z) are precisely 2, 3, 4, 5, 6, 8, 10 and 12. Since this is computational, we leave the details to the reader. In this section, we explicitly identify matrices in Sp(4, Z) of these orders. The main tool we use is Bender’s presentation of Sp(4, Z). Throughout this section we use the same notation as in [1] in all our computations. In [1], Bender gives a presentation of Sp(4, Z) using two generators and eight defining relations. We recall his result below. 4 Theorem 4.1 (Bender). The group Sp(4, Z) is generated by the two ele- ments 1 0 0 0 0 0 −1 0 1 −1 0 0 0 0 0 −1 K = , L = 0 0 1 1 10 1 0 0 0 0 −1 01 0 0 subject to the following eight relations: a) K2 = I, b) L12 = I, c) (KL7KL5K)L = L(KL5KL7K), d) (L2KL4)(KL5KL7K) = (KL5KL7K)(L2KL4), e) (L3KL3)(KL5KL7K) = (KL5KL7K)(L3KL3), f) (L2(KL5KL7K))2 = ((KL5KL7K)L2)2, g) L(L6(KL5KL7K))2 = (L6(KL5KL7K))2L, h) (KL5)5 = (L6(KL5KL7K))2.